# Real analysis limit problem.

1. Dec 2, 2014

### magimag

1. The problem statement, all variables and given/known data
Let a1=0, a2=1, and a(n+2)=n*a(n+1)+an/n+1

a)Calculate the value of a6 and a7
b)Prove that (an) converges.
c)Show that lim an=1-e-1 when n goes to infinity.

3. The attempt at a solution
I got the a part and found out that a6 19/30 and a7)91/144

part b)
each subsequent term becomes less in value, and therefore it is converging. How can prove that it's a Cauchy sequence?

part c)
I'm lost here.

Any hints are appreciated :) I'm lost on this one.

2. Dec 3, 2014

### RUber

Did you forget a parenthesis? What you wrote above does not converge.
$a_1=0, a_2 = 1, a_3 = 1*1+0/2 = 1, a_4 = 2*1+1/3 = 7/3, a_5 = 3*7/3+1/4 =29/4...$
or did you mean:
$a_1=0, a_2 = 1, a_3 = (1*1+0)/2 = 1/2, a_4 = (2*1/2+1)/3 = 2/3, a_5 = (3*2/3+1/2)/4 =5/8, a_6 = (4*5/8 +2/3)/5=19/30?$
Since the latter a_6 matches your post, I will assume that is correct.

I would start by trying some of your basic convergence tests...see if you can pick out some patterns.
Note that if $a_{n+2}=\frac{n a_{n+1}+a_n}{n+1}$, then $a_{n+3}=\frac{(n+1) a_{n+2}+a_{n+1}}{n+2}=\frac{(n+1) \left(\frac{n a_{n+1}+a_n}{n+1}\right)+a_{n+1}}{n+2}$.

3. Dec 3, 2014

### pasmith

This is false: you have $\min(a_n,a_{n+1}) \leq a_{n+2} \leq \max(a_n,a_{n+1})$ so the sequence is neither increasing nor decreasing.

First, show that $$a_{n+2} - a_{n+1} = \frac{-1}{n+1}(a_{n+1} - a_n)$$ and hence that $$a_{n+1} - a_n = \frac{-(-1)^n}{n!}.$$ If $n \geq 3$ then $$|a_{n+1} - a_n| = \frac{1}{n!} < \frac{1}{2^{n-1}}.$$
It is straightforward to show that a sequence $b_n$ which satisfies $$|b_{n+1} - b_n| < \frac{1}{2^{n-1}}$$ for all $n \geq N_0 \in \mathbb{N}$ is cauchy.

I suggested that you prove that $$a_{k+1} - a_k = -\frac{(-1)^k}{k!}.$$ You can then find an expression for $a_n$ from $$\sum_{k=1}^{n-1} (a_{k+1} - a_k) = -\sum_{k=1}^{n-1} \frac{(-1)^k}{k!}.$$